Staff: Mentor

Hello @pasmith, I dont understand;
On WHAT basis and WHY are you assuming [itex]|x - a| < 1[/itex]??

Amad, be careful when you quote someone. @pasmith did not write "|x−a|<1|x - a| < 1" - what he wrote was "If we assume |x−a| < 1 then ..."

He made this assumption, and if it turns out that this assumption is incorrect, all that is accounted for in the last line of his post where he says that |s - a| will be less than the minimum of 1 and the expression involving ##\epsilon##.

He made this assumption, and if it turns out that this assumption is incorrect, all that is accounted for in the last line of his post where he says that |s - a| will be less than the minimum of 1 and the expression involving ϵ\epsilon.

The "[itex]\delta[/itex]" was derived under the assumption that |a- 1|< 1 (so that -1< a- 1< 1 and then 1< a+ 1< 3). In order for the proof to work, both [tex]|a- 1|< \delta[/tex] and [tex]|a- 1|< 1[/tex] must be true. That will be the case if it is less than the smaller of the two.

Staff: Mentor

We don't care about that. You're missing the big picture here. The goal is to see how close x needs to be to a so that |x2 - a2| < ##\epsilon##.
We don't care about any x values that are relatively distant from a.

We don't care about that. You're missing the big picture here. The goal is to see how close x needs to be to a so that |x2 - a2| < ϵ\epsilon.
We don't care about any x values that are relatively distant from a.

I really dont understand this concept.

The definition says you have to prove there is some [itex]\delta[/itex] such that [itex] |x - a| < \delta[/itex] not that [itex]|x - a| < \delta < 1[/itex]

Also, there is no proper definition of "relatively distant." Relatively close could mean 10000000000000. Why are you making assumptions anyway?

To conclude that [itex]x^2 - a^2[/itex] is continuous at [itex]a[/itex] you need to prove:

Statement 1
For all [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that for all [itex]x[/itex], if [itex]|x - a| < \delta[/itex] then [itex]|x^2 - a^2| < \epsilon[/itex].

What Spivak invites you to prove is a stronger statement:

Statement 2
For all [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that [itex]\delta < 1[/itex] and for all [itex]x[/itex], if [itex]|x - a| < \delta[/itex] then [itex]|x^2 - a^2| < \epsilon[/itex].

Note that Statement 2 implies Statement 1. Of course one could substitute any [itex]R > 0[/itex] for '1' in Statement 2 and the result would still imply Statement 1, but 1 is the most natural choice.